WAEC Further Mathematics Past Questions & Answers

221.

Simplify $8^{n} \times 2^{2n} \div 4^{3n}$

$2^{-n}$

$2^{1 - n}$

$2^{n}$

$2^{n + 1}$

View answer & explanation

Correct answer is A

$8^{n} \times 2^{2n} \div 4^{3n} = 2^{3n} \times 2^{2n} \div 2^{6n}$

$2^{2n + 3n - 6n}$

= $2^{-n}$

222.

Which of the following is a singular matrix?

$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$

$\begin{pmatrix} 3 & 8 \\ 5 & 16 \end{pmatrix}$

$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

View answer & explanation

Correct answer is A

No explanation has been provided for this answer.

223.

Simplify $\frac{^{n}P_{4}}{^{n}C_{4}}$

View answer & explanation

Correct answer is A

$^{n}P_{4} = \frac{n!}{(n - 4)!}$

$^{n}C_{4} = \frac{n!}{(n - 4)! 4!}$

$\frac{^{n}P_{4}}{^{n}C_{4}} = \frac{n!}{(n - 4)!} ÷ \frac{n!}{(n - 4)! 4!}$

= $4! = 24$

224.

A committee of 4 is to be selected from a group of 5 men and 3 women. In how many ways can this be done if the chairman of the committee must be a man?

175

View answer & explanation

Correct answer is D

From the five(5) men viable for the position of the chairman, one of them must be selected.

This implies; 4 men and 3 women are open for committee membership.

7 C $_3$ = $\frac{7!}{3!(7 - 3)!}$ =

$\frac{7⋅6⋅5⋅4⋅3⋅2⋅1}{3.2.1 \times 4.3.2.1}$

= 35 $\times$ 5 = 175

225.

The mean age of n men in a club is 50 years. Two men aged 55 and 63 years left the club, and the mean age reduced by 1 year. Find the value of n.

View answer & explanation

Correct answer is B

Let the sum of the men's ages be f, so that

$\frac{f}{n} = 50 .... (1)$

Also, $\frac{f - (55 + 63)}{n - 2} = 50 - 1 = 49 .... (2)$

From (1), $f = 50n$

From (2), $f - 118 = 49(n - 2) = 49n - 98$

$f = 49n - 98 + 118 = 49n + 20$

$\therefore f = 50n = 49n + 20$

$50n - 49n = n = 20$

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